Re: Is this possible? Residue computation leads to complexInfinity?

*To*: mathgroup at smc.vnet.net*Subject*: [mg60725] Re: [mg60719] Is this possible? Residue computation leads to complexInfinity?*From*: Bob Hanlon <hanlonr at cox.net>*Date*: Mon, 26 Sep 2005 01:36:04 -0400 (EDT)*Reply-to*: hanlonr at cox.net*Sender*: owner-wri-mathgroup at wolfram.com

Your Residue expression is unevaluated (Residue appears in output) since you left out part of the Residue syntax. However, your argument was evaluated to ComplexInfinity so z must already have had a value. Clear[z]; FullSimplify[ Residue[(I*z)/(1-(2+4*a)*z^2+z^4), {z,Sqrt[a+1]-Sqrt[a]}],a>0] (I*(Sqrt[1 + 1/a] - 1))/(8*(-a + Sqrt[a^2 + a] - 1)) Bob Hanlon > > From: "kiki" <lunaliu3 at yahoo.com> To: mathgroup at smc.vnet.net > Date: 2005/09/25 Sun AM 02:36:33 EDT > Subject: [mg60725] [mg60719] Is this possible? Residue computation leads to complexInfinity? > > In[15]:= > \!\(FullSimplify[Residue[\(\[ImaginaryI]\ z\)\/\(1 - \((2 + 4\ a)\)\ z\^2 + > z\ > \^4\), \@\(a + 1\) - \@a], \ a > 0]\) > > Out[15]= > \!\(Residue[ComplexInfinity, \(-\@a\) + \@\(1 + a\)]\) > > >